//给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。 
//
// 岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。 
//
// 此外，你可以假设该网格的四条边均被水包围。 
//
// 
//
// 示例 1： 
//
// 
//输入：grid = [
//  ["1","1","1","1","0"],
//  ["1","1","0","1","0"],
//  ["1","1","0","0","0"],
//  ["0","0","0","0","0"]
//]
//输出：1
// 
//
// 示例 2： 
//
// 
//输入：grid = [
//  ["1","1","0","0","0"],
//  ["1","1","0","0","0"],
//  ["0","0","1","0","0"],
//  ["0","0","0","1","1"]
//]
//输出：3
// 
//
// 
//
// 提示： 
//
// 
// m == grid.length 
// n == grid[i].length 
// 1 <= m, n <= 300 
// grid[i][j] 的值为 '0' 或 '1' 
// 
// 👍 1384 👎 0


package leetcode.editor.cn;

//岛屿数量

import 分类.DFS;
import leetcode.editor.cn.level.中等;

public class P200_岛屿数量 implements DFS, 中等 {
    public static void main(String[] args) {
        Solution solution = new P200_岛屿数量().new Solution();

    }

    //leetcode submit region begin(Prohibit modification and deletion)
    class Solution {
        public int numIslands(char[][] grid) {
            int count = 0;
            for (int i = 0; i < grid.length; i++) {
                for (int j = 0; j < grid[i].length; j++) {
                    // 遇到陆地后累计加1  然后将其淹没
                    if (grid[i][j] == '1') {
                        count++;
                        dfs(grid, i, j);
                    }
                }
            }             
            return count;
        }
        
        private void dfs(char[][] grid, int x, int y) {
            if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length) {
                return;
            }
            // 遇到海水就返回
            if (grid[x][y] == '0') {
                return;
            }
            // 淹掉陆地
            grid[x][y] = '0';
            dfs(grid, x-1, y);
            dfs(grid, x+1, y);
            dfs(grid, x, y-1);
            dfs(grid, x, y+1);
        }
    }
//leetcode submit region end(Prohibit modification and deletion)


}